Problem: Simplify and expand the following expression: $ \dfrac{1}{2r + 12}- \dfrac{5}{r + 8}- \dfrac{2}{r^2 + 14r + 48} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the first term: $ \dfrac{1}{2r + 12} = \dfrac{1}{2(r + 6)}$ We can factor the quadratic in the third term: $ \dfrac{2}{r^2 + 14r + 48} = \dfrac{2}{(r + 6)(r + 8)}$ Now we have: $ \dfrac{1}{2(r + 6)}- \dfrac{5}{r + 8}- \dfrac{2}{(r + 6)(r + 8)} $ The least common multiple of the denominators is: $ 2(r + 6)(r + 8)$ In order to get the first term over $2(r + 6)(r + 8)$ , multiply by $\dfrac{r + 8}{r + 8}$ $ \dfrac{1}{2(r + 6)} \times \dfrac{r + 8}{r + 8} = \dfrac{r + 8}{2(r + 6)(r + 8)} $ In order to get the second term over $2(r + 6)(r + 8)$ , multiply by $\dfrac{2(r + 6)}{2(r + 6)}$ $ \dfrac{5}{r + 8} \times \dfrac{2(r + 6)}{2(r + 6)} = \dfrac{10(r + 6)}{2(r + 6)(r + 8)} $ In order to get the third term over $2(r + 6)(r + 8)$ , multiply by $\dfrac{2}{2}$ $ \dfrac{2}{(r + 6)(r + 8)} \times \dfrac{2}{2} = \dfrac{4}{2(r + 6)(r + 8)} $ Now we have: $ \dfrac{r + 8}{2(r + 6)(r + 8)} - \dfrac{10(r + 6)}{2(r + 6)(r + 8)} - \dfrac{4}{2(r + 6)(r + 8)} $ $ = \dfrac{ r + 8 - 10(r + 6) - 4} {2(r + 6)(r + 8)} $ Expand: $ = \dfrac{r + 8 - 10r - 60 - 4}{2r^2 + 28r + 96} $ $ = \dfrac{-9r - 56}{2r^2 + 28r + 96}$